Question

If $x$ and $y$ are real numbers which satisfy the relation $x^2+9y^2-4x+6y+4=0,$ then the maximum value of $\frac{(4x-9y)}{2}$ is

Answer 1

$x^2+9y^2-4x+6y+4=0$
$\Rightarrow x^2-4x+4+9y^2+6y+1=1$
$\Rightarrow (x-2{)}^2+(3y+1{)}^2=1$
$\Rightarrow (x-2{)}^2+\frac{{(y+\frac{1}{3})}^2}{\frac{1}{9}}=1$
which is an equation of ellipse having centre at $(2,-\frac{1}{3}).$

If $P(x,y)$ is any point on ellipse, then
$x=2+\cos \theta$ and $y=-\frac{1}{3}+\frac{1}{3}\sin \theta$

$\therefore 4x-9y=4(2+\cos \theta )-9(-\frac{1}{3}+\frac{1}{3}\sin \theta )$
$\Rightarrow f\left(\theta \right)=8+4\cos \theta +3-3\sin \theta$
$\Rightarrow f\left(\theta \right)=11+4\cos \theta -3\sin \theta$
$\Rightarrow f(\theta {)}_{max}=11+\sqrt{3^2+4^2}=16$
So maximum value of $\frac{4x-9y}{2}=\frac{16}{2}=8$